Search any question & find its solution
Question:
Answered & Verified by Expert
A body starts from rest, under the action of an engine working at a constant power and moves along a straight line. The displacement s is given as a function of time ( $f$ as
Options:
Solution:
1181 Upvotes
Verified Answer
The correct answer is:
$s=a t^{3 / 2},$ a is a constant
Given, Power $(P)=$ constant
Kinetic Energy (KE) $=\frac{1}{2} m v^{2}$
We know that, $P=\frac{K E}{\Delta t} \Rightarrow P=\frac{m v^{2}}{\Delta t}$
$\because P=$ constant.
Hence, velocity of the body $v \propto \sqrt{t}$...(i)
As, Velocity $v=\frac{d s}{d t}$...(ii)
From Eqs. (i) and (ii), we get
So, $\quad \frac{d s}{d t} \propto \sqrt{t}$
Integrating the above cquation w.r.t. time $(t)$
$$
\int \frac{d s}{d t} \propto \int \sqrt{t}
$$
we get, displacement of the body $s \propto t^{3 / 2}$
Displacement $s=a t^{3 / 2}$, where $a$ is constant.
Kinetic Energy (KE) $=\frac{1}{2} m v^{2}$
We know that, $P=\frac{K E}{\Delta t} \Rightarrow P=\frac{m v^{2}}{\Delta t}$
$\because P=$ constant.
Hence, velocity of the body $v \propto \sqrt{t}$...(i)
As, Velocity $v=\frac{d s}{d t}$...(ii)
From Eqs. (i) and (ii), we get
So, $\quad \frac{d s}{d t} \propto \sqrt{t}$
Integrating the above cquation w.r.t. time $(t)$
$$
\int \frac{d s}{d t} \propto \int \sqrt{t}
$$
we get, displacement of the body $s \propto t^{3 / 2}$
Displacement $s=a t^{3 / 2}$, where $a$ is constant.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.